[proofplan]
The identity at time zero is part of the definition of the local flow as the solution map for the [initial value problem](/page/Initial%20Value%20Problem). For the composition law, we fix an intermediate time $s$ and compare two curves of the new time variable $r$: the solution started from $\varphi(s,x)$ and the time-shift of the solution started from $x$. Both curves solve the same autonomous ODE and have the same value at $r=0$, so uniqueness for locally Lipschitz ODEs forces them to agree on their common interval of definition. Evaluating that agreement at the prescribed time $t$ gives the asserted local flow identity.
[/proofplan]
[step:Read the identity at time zero from the definition of the local flow]
For each $x\in E$, the local flow convention says that $J_x=\{r\in\mathbb{R}:(r,x)\in\mathcal{D}\}$ is the maximal interval of the solution of the initial value problem
\begin{align*}
\dot{z}(r)=V(z(r)),\qquad z(0)=x.
\end{align*}
Hence $0\in J_x$ and
\begin{align*}
\varphi(0,x)=x.
\end{align*}
[/step]
[step:Compare the shifted trajectory with the trajectory from the intermediate point]
Fix $x\in E$ and $s,t\in\mathbb{R}$ such that $(s,x)\in\mathcal{D}$, $(t,\varphi(s,x))\in\mathcal{D}$, and $(t+s,x)\in\mathcal{D}$. Define
\begin{align*}
y:=\varphi(s,x)\in E.
\end{align*}
Let
\begin{align*}
J_y:=\{r\in\mathbb{R}:(r,y)\in\mathcal{D}\}
\end{align*}
be the maximal interval for the solution through $y$. Also define the shifted interval
\begin{align*}
K:=\{r\in\mathbb{R}:r+s\in J_x\}.
\end{align*}
Since $s\in J_x$, the interval $K$ contains $0$. Define the shifted trajectory $v:K\to E$ by
\begin{align*}
v(r):=\varphi(r+s,x).
\end{align*}
The map $r\mapsto \varphi(r+s,x)$ is differentiable on $K$, and by the chain rule and the differential equation for the trajectory through $x$,
\begin{align*}
\dot{v}(r)=V(\varphi(r+s,x))=V(v(r))
\end{align*}
for every $r\in K$. Its initial value is
\begin{align*}
v(0)=\varphi(s,x)=y.
\end{align*}
Thus $v$ is a solution of $\dot{z}=V(z)$ on $K$ with initial value $y$ at time $0$.
[guided]
We fix the starting point $x$ and the intermediate time $s$, and we set
\begin{align*}
y:=\varphi(s,x).
\end{align*}
The goal is to show that evolving first from $x$ for time $s$ and then from $y$ for time $t$ gives the same point as evolving directly from $x$ for time $s+t$.
The solution through $y$ is defined on
\begin{align*}
J_y:=\{r\in\mathbb{R}:(r,y)\in\mathcal{D}\}.
\end{align*}
To compare it with the original solution through $x$, we re-center the original trajectory at time $s$. Define
\begin{align*}
K:=\{r\in\mathbb{R}:r+s\in J_x\}.
\end{align*}
Because $(s,x)\in\mathcal{D}$, we have $s\in J_x$, so $0\in K$. Now define the shifted trajectory $v:K\to E$ by
\begin{align*}
v(r):=\varphi(r+s,x).
\end{align*}
This is the right comparison curve because at the new time $r=0$ it starts exactly at the intermediate point:
\begin{align*}
v(0)=\varphi(s,x)=y.
\end{align*}
It remains to check that it satisfies the same differential equation as the solution through $y$. Since the original trajectory $q:J_x\to E$ given by $q(\tau)=\varphi(\tau,x)$ solves $\dot{q}(\tau)=V(q(\tau))$, the chain rule applied to $v(r)=q(r+s)$ gives
\begin{align*}
\dot{v}(r)=\dot{q}(r+s)=V(q(r+s))=V(\varphi(r+s,x))=V(v(r)).
\end{align*}
Therefore $v$ is a solution of $\dot{z}=V(z)$ on the interval $K$ with initial value $v(0)=y$.
[/guided]
[/step]
[step:Use uniqueness to identify the two trajectories on their common domain]
Define $u:J_y\to E$ by
\begin{align*}
u(r):=\varphi(r,y).
\end{align*}
By the definition of the flow, $u$ is the maximal solution of $\dot{z}=V(z)$ with $u(0)=y$. The vector field $V:E\to\mathbb{R}^n$ is locally Lipschitz, hence continuous, so the autonomous map $F:\mathbb{R}\times E\to\mathbb{R}^n$ defined by $F(r,z)=V(z)$ satisfies the hypotheses of [citetheorem:8375]. Applying uniqueness from that theorem to the two solutions $u:J_y\to E$ and $v:K\to E$ with the same initial value $y$ gives
\begin{align*}
u(r)=v(r)
\end{align*}
for every $r\in J_y\cap K$.
The hypotheses $(t,\varphi(s,x))\in\mathcal{D}$ and $(t+s,x)\in\mathcal{D}$ say exactly that $t\in J_y$ and $t\in K$. Therefore $t\in J_y\cap K$, and substituting $r=t$ yields
\begin{align*}
\varphi(t,\varphi(s,x))=u(t)=v(t)=\varphi(t+s,x).
\end{align*}
This proves the local flow property.
[/step]
